# How do you differentiate g(y) =(y^-1 + y^-2)(5y^2 -y^3) using the product rule?

Jul 2, 2016

$g ' \left(y\right) = 4 - 2 y$.

#### Explanation:

$g \left(y\right) = \left({y}^{-} 1 + {y}^{-} 2\right) \left(5 {y}^{2} - {y}^{3}\right) = \left({y}^{-} 1 + {y}^{-} 2\right) {y}^{2} \left(5 - y\right) = \left({y}^{-} 1 \cdot {y}^{2} + {y}^{-} 2 \cdot {y}^{2}\right) \left(5 - y\right) = \left(y + 1\right) \left(5 - y\right)$

Using Product Rule $: \frac{d}{\mathrm{dy}} \left(U \cdot V\right) = U \cdot \frac{d}{\mathrm{dy}} V + V \cdot \frac{d}{\mathrm{dy}} U$

$g ' \left(y\right) = \left(y + 1\right) \frac{d}{\mathrm{dy}} \left(5 - y\right) + \left(5 - y\right) \frac{d}{\mathrm{dy}} \left(y + 1\right)$

$= \left(y + 1\right) \left\{\frac{d}{\mathrm{dy}} 5 - \frac{d}{\mathrm{dy}} y\right\} + \left(5 - y\right) \left\{\frac{d}{\mathrm{dy}} y + \frac{d}{\mathrm{dy}} 1\right\}$

$= \left(y + 1\right) \left(0 - 1\right) + \left(5 - y\right) \left(1 + 0\right) = - y - 1 + 5 - y = 4 - 2 y$

Note that we were reqd. to diif. using Product Rule. Otherwise,

$g \left(y\right) = \left(y + 1\right) \left(5 - y\right) = 5 y + 5 - {y}^{2} - y = 5 + 4 y - {y}^{2}$' so that,

$g ' \left(y\right) = \left(5\right) ' + \left(4 y\right) ' - \left({y}^{2}\right) ' = 0 + 4 \cdot 1 - 2 y = 4 - 2 y$, as before!

I hope this will be of help! Enjoy Maths.!